15 thoughts on “Year 6 Maths Challenge

  1. A is 4,B is 8, C is 9,D is 1,E is 3,F is 6
    And G is 5.

    1) A+A (4+4=8) A4
    2) AxA=DF(4×4=16) B8
    3) A+C=DE(4+9=19) C9
    4) C+C=DB(9+9=18) D1
    5) CxC=BD(9×9=81) E3
    6) AxC=EF(4×9=36) F6

  2. A is 4, B is 8, C is 9, D is 1, E is 3, F is 6 and
    G is 5.

    1. A+A=B which is 4+4=8

    2. AxA=DF which is 4×4=16

    3. A+C=DE which is 4+9=13

    4. C+C=DB which is 9+9=18

    5. CxC=BD which is 9×9=81

    6. AxC=EF which is 4×9=36

    I had to do trial and error and try different possibilities to get the answer.

  3. A=4
    A+A=B (4+4=8)
    AxA=DF (4×4=16)
    AxC=EF (4×9=36)
    C+C=DB (9+9=18)
    CxC=BD (9×9=81)
    AxC=EF (4×9=36)

    To crack the code,I got the letters that had the least amount of possibilities.That was A+A=B.Then,I found all the possibilities and tried the other codes with the possibilities.To find the all the possibilities for A+A=B I had to work systematically.

  4. For this question i will be using a systematic approach to find my answer. This will help me because I can go back and check my answers.

    A+A=B (4+4=8)
    AxA=DF (4×4=16)
    AxC=EF (4×9=36)
    C+C=DB (9+9=18)
    CxC=BD (9×9=81)
    AxC=EF (4×9=36)
    A=4 B=8 C=9 D=1 E=3 F=6

    Working out is provided in homework book.

  5. There is 7 letters that need numbers so it is going to be trial and error .


    A+A=B [4+4=8]
    A*A=DF [4*4=16]
    A+C=DE [4+9+13]
    C+C=DB [9+9=18]
    C*C=BD [9*9=81]
    A*C=EF [4*9=36]

    I did trial and error on a piece of paper at home an got this answer

  6. 1.A+A=B (4+4=8)
    2.C+C=DB (9+9=18)
    3.A+A=DF (4X4=16)
    4.CXC=BD (9X9=81)
    5.A+C=DE (4+9=13)
    6.AXC=EF (4X9=36) A=4 B=8 C=9 D=1 E=3 F=6

    I think I’ve answered all the questions. There are 6 letters, but there are 7 numbers. There’s no G.

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